Lösung 2.2:4c

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The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
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{{Displayed math||<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}
We take out a factor 4 from the denominator
We take out a factor 4 from the denominator
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{{Displayed math||<math>\int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}</math>}}
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{{Abgesetzte Formel||<math>\int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}</math>}}
and rewrite the quadratic term as
and rewrite the quadratic term as
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{{Displayed math||<math>\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}</math>}}
If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
&= \left\{\begin{align}
&= \left\{\begin{align}

Version vom 13:03, 10. Mär. 2009

The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,

\displaystyle \int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}

We take out a factor 4 from the denominator

\displaystyle \int \frac{dx}{(x+2)^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}(x+2)^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1}

and rewrite the quadratic term as

\displaystyle \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}(x+2)^2+1} = \frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}\,\textrm{.}

If we now substitute \displaystyle u = (x+2)/2, we obtain the integral in the exercise

\displaystyle \begin{align}

\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1} &= \left\{\begin{align} u &= (x+2)/2\\[5pt] du &= dx/2 \end{align}\right\}\\[5pt] &= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt] &= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt] &= \frac{1}{2}\arctan u + C\\[5pt] &= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.} \end{align}

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